Holomorphic Function

In complex analysis, a holomorphic function is a complex differentiable function. The condition of complex differentiability is very strong, and leads to an especially elegant theory of calculus for these functions.

A function $f: \mathbb{R} \to \mathbb{R}$ is said to be differentiable at $x\in \mathbb{R}$ if $$\lim_{\overset{h\in \mathbb{R}}{h\to 0}} \frac{f(x+h) - f(x)}{h}$$ exists. If the limit exists, its value is called $f'(x)$.

Analogously, a function $f: \mathbb{C} \to \mathbb{C}$ $($that is, $f$ takes a complex number and returns a complex number$)$ is said to be complex differentiable at $z\in \mathbb{C}$ if $$\lim_{\overset{h\in \mathbb{C}}{h\to 0}} \frac{f(z+h) - f(z)}{h}$$ exists. Again, if the limit exists, its value is called $f'(z)$. If $f$ is complex differentiable at every $z\in U \subset \mathbb{C}$, then $f$ is said to be holomorphic on $U$.

Although both conditions are notationally the same, the complex differentiability condition is actually much stronger than the differentiability condition for functions $\mathbb{R} \to \mathbb{R}$.

This is because as $h$ approaches $0$ in the complex limit, it must do so from all directions, of which there are infinitely many. (In the real limit, there are only two directions from which $h$ may approach $0$: the positive direction and the negative direction.) Thus, if $f: \mathbb{C} \to \mathbb{C}$ is complex differentiable at $z\in \mathbb{C}$, then $$\lim_{\overset{t\in \mathbb{R}}{t\to 0}} \frac{f(z+th) - f(z)}{th}$$ exists for all $h\in \mathbb{C}$ and equals $f'(z)$.

One can view $\mathbb{C}$ as $\mathbb{R}^2$ by identifying the complex number $x+yi$ with the pair $(x,y)$. From this perspective, one may write $f(z) = f(x,y) = u(x,y) + i \cdot v(x,y)$, where $u(x,y) = \text{Re}\big(f(z)\big)$ and $v(x,y) = \text{Im}\big(f(z)\big)$. But then the complex differentiability condition requires $$\begin{aligned} f'(z) &= \lim_{\overset{t\in \mathbb{R}}{ t\to 0}} \frac{f(z+t \cdot 1) - f(z)}{t\cdot 1} \\ &= \lim_{\overset{t\in \mathbb{R}}{ t\to 0}} \left(\frac{u(x+t,y) - u(x,y)}{t} + i \cdot \frac{v(x+t,y) - v(x,y)}{t} \right)\\ &= \left. \frac{\partial u}{\partial x}\right|_{(x,y)} + i \left. \frac{\partial v}{\partial x} \right|_{(x,y)} \end{aligned}$$ and $$\begin{aligned} f'(z) &= \lim_{\overset{t\in \mathbb{R}}{ t\to 0}} \frac{f(z+t \cdot i) - f(z)}{t\cdot i} \\ &= -i \cdot \lim_{\overset{t\in \mathbb{R}}{ t\to 0}} \left(\frac{u(x,y+t) - u(x,y)}{t} + i \cdot \frac{v(x,y+t) - v(x,y)}{t} \right)\\ &= -i \cdot \left( \frac{\partial u}{\partial y} + i \cdot \frac{\partial v}{\partial y} \right) \\ &=\left. \frac{\partial v}{\partial y}\right|_{(x,y)} - i \cdot \left. \frac{\partial u}{\partial y} \right|_{(x,y)}. \end{aligned}$$ Setting these equal proves that wherever $f$ is complex differentiable, it satisfies $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\quad \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}.$$ These are the Cauchy-Riemann equations, a system of differential equations that every holomorphic function must satisfy on its domain. Hence, there are severe constraints on how a holomorphic function may behave; namely, it may only behave as these equations prescribe.

Perhaps the most important property of holomorphic functions is their analyticity. This means that any holomorphic function has an expansion as a power series.

Suppose $f: U \to \mathbb{C}$ is holomorphic with $U$ an open set. Then there exists a sequence of complex numbers $\{a_n\}_{n\ge 0}$ such that $$f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n$$ within the radius of convergence about $z_0$ (i.e., it converges for all $z$ in some disk centered at $z_0$) and equals $f(z)$ when it converges. Specifically, $a_n = \frac{f^{(n)}(z_0)}{n!}$.

Let $\gamma$ be a closed curve. Using the Cauchy integral formula and geometric series, we have

$$\begin{aligned} f(z) &= \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)}{\zeta-z} \ d\zeta \\ &= \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)/(\zeta-z_0)}{1-\frac{z-z_0}{\zeta-z_0}} \ d\zeta \\ &= \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)}{\zeta-z_0} \sum_{n=0}^\infty \left( \frac{z-z_0}{\zeta-z_0} \right)^n \ d\zeta \\ &= \sum_{n=0}^\infty \left( \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)}{(\zeta-z_0)^{n+1}} \ d\zeta \right) (z-z_0)^n \\ &= \sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n.\ _\square \end{aligned}$$